Let $(a,b) \subseteq \mathbb R$ and $\{\varphi_n \}_{n \geq 1}$ be an orthonormal basis for $L^2((a,b)).$ Define $\varphi_{m,n} : (a,b) \times (a,b) \longrightarrow \mathbb C$ by $$\varphi_{m,n} (s,t) = \varphi_m (s) \varphi_n (t),\ s,t \in (a,b).$$ Then $\varphi_{m,n} \in L^2(a,b)$ for all $m,n \geq 1$ and $\{\varphi_{m,n} \}_{m \geq 1, n \geq 1}$ is an orthonormal basis for $L^2((a,b) \times (a,b)).$
My Attempt $:$
It is easy to show that $\{\varphi_{m,n} \}_{m \geq 1, n \geq 1}$ is an orthonormal set in $L^2(a,b).$ In order to show that it's an orthonormal basis of $L^2((a,b) \times (a,b)),$ I took a function $f \in L^2((a,b) \times (a,b))$ such that for all $m,n \geq 1$ we have $$\tag{1} \int_{a}^{b} \int_{a}^{b} f(s,t) \varphi_{m,n} (s,t)\ ds\ dt = 0.$$ We need to show that $f = 0$ almost everywhere. By Hölder's inequality we have $$\begin{align*}\int_{a}^{b} \left \lvert \int_{a}^{b} f(s,t) \varphi_m (s)\ ds \right \rvert^2\ dt & \leq \|\varphi_m \|_2^2 \int_{a}^{b} \int_{a}^{b} \left \lvert f(s,t) \right \rvert^2\ ds\ dt \\ & = \int_{a}^{b} \left \lvert f(s,t) \right\rvert^2\ ds\ dt \lt + \infty. \end{align*}$$ Thus the function $t \mapsto \displaystyle \int_{a}^{b} f(s,t) \varphi_m (s)\ ds \in L^2(a,b).$ So, by $(1)$ we have $$\tag{2}\int_{a}^{b} f(s,t) \varphi_m (s)\ ds = 0$$ for a.e. $t \in (a,b).$ Now since $f \in L^2(a,b)$ by Fubini's theorem it follows that the function $f^t \in L^2(a,b)$ for a.e. $t \in (a,b), $ where $f^t : s \mapsto f(s,t).$ Therefore by $(2),$ it implies that for a.e. $t \in (a,b),$ we have $f(s,t) = 0$ for a.e. $s \in (a,b).$
Question $:$ Does this imply that $f = 0$ a.e. on $(a,b) \times (a,b)\ $?
I am confused at this stage. Could anyone please give me suggestion regarding this one?
Thanks a bunch!
Apply Fubini/Tonelli's Theorem to $g(s,t)=\chi_{\{(s,t): f(s,t) \neq 0\}}$. Since $\int g(s,t) ds=0$ for almost all $t$, we get $\int \int g(s,t) dsdt=0$ by integration w.r.t. $t$.